cp's OEIS Frontend

Continued fraction for sqrt(3).

Also coefficient of the highest power of q in the expansion of the polynomial nu(n) defined by: nu(0)=1, nu(1)=b and for n>=2, nu(n)=b*nu(n-1)+lambda*(n-1)_q*nu(n-2) with (b,lambda)=(1,1), where (n)_q=(1+q+...+q^(n-1)) and q is a root of unity. - Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002

nu(0)=1 nu(1)=1; nu(2)=2; nu(3)=3+q; nu(4)=5+3q+2q^2; nu(5)=8+7q+6q^2+4q^3+q^4; nu(6)=13+15q+16q^2+14q^3+11q^4+5q^5+2q^6.

From Jaroslav Krizek, May 28 2010: (Start)

a(n) = denominators of arithmetic means of the first n positive integers for n >= 1.

See A026741(n+1) or A145051(n) - denominators of arithmetic means of the first n positive integers. (End)

From R. J. Mathar, Feb 16 2011: (Start)

This is a prototype of multiplicative sequences defined by a(p^e)=1 for odd primes p, and a(2^e)=c with some constant c, here c=2. They have Dirichlet generating functions (1+(c-1)/2^s)*zeta(s).

Examples are A153284, A176040 (c=3), A040005 (c=4), A021070, A176260 (c=5), A040011, A176355 (c=6), A176415 (c=7), A040019, A021059 (c=8), A040029 (c=10), A040041 (c=12). (End)

a(n) = p(-1) where p(x) is the unique degree-n polynomial such that p(k) = A000325(k) for k = 0, 1, ..., n. - Michael Somos, May 12 2012

For n > 0: denominators of row sums of the triangular enumeration of rational numbers A226314(n,k) / A054531(n,k), 1 <= k <= n; see A226555 for numerators. - Reinhard Zumkeller, Jun 10 2013

From Jianing Song, Nov 01 2022: (Start)

For n > 0, a(n) is the minimal gap of distinct numbers coprime to n. Proof: denote the minimal gap by b(n). For odd n we have A058026(n) > 0, hence b(n) = 1. For even n, since 1 and -1 are both coprime to n we have b(n) <= 2, and that b(n) >= 2 is obvious.

The maximal gap is given by A048669. (End)

Complement of A014601. - Reinhard Zumkeller, Oct 04 2004

Let S(x) = (1, 2, 2, 2, ...). Then A042963 = ((S(x))^2 + S(x^2))/2 = ((1, 4, 8, 12, 16, 20, ...) + (1, 0, 2, 0, 2, 0, 2, ...))/2 = (1, 2, 5, 6, 9, 10, ...). - Gary W. Adamson, Jan 03 2011

(a(n)*(a(n) + 1 + 4*k))/2 is odd, for k >= 0. - Gionata Neri, Jul 19 2015

Equivalent to the following variation on Fermat's Diophantine m-tuple: 1 + the product of any two distinct terms is not a square; this sequence, which we'll call sequence S, is produced by the following algorithm. At the start, S is initially empty. At stage n, starting at n = 1, the algorithm checks whether there exists a number m already in the sequence, such that mn+1 is a perfect square. If such a number m is found, then n is not added to the sequence; otherwise, n is added. Then n is incremented to n + 1, and we repeat the procedure. Proof by Clark R. Lyons: We prove by strong induction that n is in the sequence S if and only if n == 1 (mod 4) or n == 2 (mod 4). Suppose now that this holds for all k < n. In case 1, either n == 1 (mod 4) or n == 2 (mod 4), and we wish to show that n does indeed enter the sequence S. That is, we wish to show that there does not exist m < n, already in the sequence at this point such that mn+1 is a square. By the inductive hypothesis m == 1 (mod 4) or m == 2 (mod 4). This means that both m and n are one of 1, 2, 5, or 6 mod 8. Using a multiplication table mod 8, we see that this implies mn+1 is congruent to one of 2, 3, 5, 6, or 7 mod 8. But we also see that mod 8, a perfect square is congruent to 0, 1, or 4. Thus mn+1 is not a perfect square, so n is added to the sequence. In case 2, n == 0 (mod 4) or n == 3 (mod 4), and we wish to show that n is not added to the sequence. That is, we wish to show that there exists m < n already in the sequence such that mn+1 is a perfect square. For this we let m = n - 2, which is positive since n >= 3. By the inductive hypothesis, since m == 1 (mod 4) or m == 2 (mod 4) and m < n, m is already in the sequence. And we have m*n + 1 = (n - 2)*n + 1 = n^2 - 2*n + 1 = (n - 1)^2, so mn+1 is indeed a perfect square, and so n is not added to the sequence. Thus n is added to the sequence if and only if n == 1 (mod 4) or n == 2 (mod 4). This completes the proof. - Robert C. Lyons, Jun 30 2016

Also the number of maximal cliques in the (n + 1) X (n + 1) black bishop graph. - Eric W. Weisstein, Dec 01 2017

Lexicographically earliest sequence of distinct positive integers such that the average of any two or more consecutive terms is never an integer. (For opposite property see A005408.) - Ivan Neretin, Dec 21 2017

Numbers whose binary reflected Gray code (A014550) ends with 1. - Amiram Eldar, May 17 2021

Also: append its negated last bit to n-1. - M. F. Hasler, Oct 17 2022

Interleaving of A010701 and A000012.

Also continued fraction expansion of (3+sqrt(21))/2.

Also decimal expansion of 31/99.

Essentially first differences of A014601.

Inverse binomial transform of 3 followed by A020707.

Second inverse binomial transform of A052919 without initial term 2.

Third inverse binomial transform of A007582 without initial term 1.

Exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 2*x^2 + 2*x^3 + 3*x^4 + 3*x^5 + ... is the o.g.f. for A008619. - Peter Bala, Mar 13 2015

As an infinite lower triangular matrix M; M * [1,2,3,...] = A063210: (1, 2, 6, 6, 10, 10, 14, 14, ...).

M * [1, 3, 5, 7, ...] = A047471, {1,3} mod 8.

Eigensequence of the triangle = A066983 starting (1, 1, 3, 3, 7, 9, 17, 25, ...).

Binomial transform of the triangle = A153861.

Row sums = A153284: (1, 1, 3, 1, 3, 1, 3, 1, ...).

1 followed by interleaving of A004767 and A017653. - Klaus Brockhaus, Jan 04 2009

1 followed by interleaving of A154105 and 3*A154106. - Klaus Brockhaus, Jan 04 2009

In the so-called lightbulb process, on days r = 1, ..., n, out of n lightbulbs, all initially off, exactly r bulbs selected uniformly and independent of the past have their status changed from off to on, or vice versa. With W_n the number of bulbs on at the terminal time n and C_n a suitable clubbed binomial distribution, d_{TV}(W_n,C_n) <= 2.7314 sqrt{n} e^{-(n+1)/3} for all n >= 1.

This is the value of the function g_1(9) after eq (16) of the preprint.